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Show line through point that is tangent to f(x)does not exist

  1. Mar 18, 2012 #1
    1. show that there is no line through the point (2,7) that is tangent to the parabola y =x^2 +x



    2. y-y1=m(x-x1)



    3.
    [itex]y'=f'(x)=2x+1[/itex]
    [itex]m1=2x +1[/itex]
    [itex]m1=2(2) +1 =5[/itex]

    [itex]m2=((x^2 +x)-7)/(x-2)[/itex]

    [itex]m1=m2?[/itex]
    [itex]((x^2 +x)-7)/(x-2)=5[/itex]
    [itex]x^2-4x+3=0[/itex]

    [itex]2^2-4(2)+3=/=0[/itex]
    [itex]-1=0[/itex]


    I'm thinking that i would compare the slope of the line passing through (2,7) to the general slope of the parabola. The other thing i believe i could do is just plug 2 into y=x^2 +x. that would give me 6 which does not equal 7, but that does not involve using any of the material we're covering in class so that is out of the question.
     
  2. jcsd
  3. Mar 18, 2012 #2
    One way to do it is to pick a random value for our hypothetical line to be tangent to y at: let's call it x1. (If that wasn't clear, we're assuming that we have a line that is tangent to the parabola at x-coordinate x1 and that also contains (2,7). We'll try to derive a contradiction to show that this is impossible). This is basically your idea to compare the slope of a random line through (2,7) to the general parabola slope.

    Since our line is tangent to the graph of y at x1, then it has slope 2x1 + 1.
    Since our line is tangent to the graph of y at x1, then it contains the point (x1, x1^2 + x1).
    So we have a line that contains (x1,x1^2 + x1) and (2,7).
    Calculating the slope of such a line using rise/over run, we get a new slope.
    By assumption this new slope must equal the old slope, 2x1 + 1.
    From there it is easy to set up a quadratic equation and show that it has no real solutions.
    This shows that such a line cannot exist!

    Your original idea to plug 2 into y does not work because we are only looking for a line that is tangent to y and contains (2,7). The only way your idea would work is if they were asking you to show that (2,7) is not a point on the parabola y.

    I hope that is helpful!
     
  4. Mar 18, 2012 #3

    Dick

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    You are almost there. But the m1=m2 equation you want to try to solve is [itex]((x^2 +x)-7)/(x-2)=2x+1[/itex]. Don't substitute x=2 into m1. You don't know the tangent line hits the parabola at x=2. Draw a picture.
     
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